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In a recent poll of 1200 randomly selected adult office workers, 32% said they had worn a Halloween costume to the office at least once. Report the 95% confidence interval for the proportion of all adult office workers who have worn a Halloween costume to the office at least once. (Round final calculations to the nearest tenth of a percent)

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JeanaShupp

Answer: [tex](29.4\%,\ 34.6\%)[/tex]

Step-by-step explanation:

The confidence interval for population proportion is given by :-

[tex]\hat{p}\pm z*\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}[/tex]

, where [tex]\hat{p}[/tex] = sample proportion.

z* = critical value (Two-tailed)

n= Sample size ;

In the given problem , we are given

n= 1200

[tex]\hat{p}=0.32[/tex]

Also, we know that the critical value for 95 %confidence interval : z*= 1.96

Now, the 95% confidence interval for the proportion of all adult office workers who have worn a Halloween costume to the office at least once will be :-

[tex]0.32\pm (1.96)\sqrt{\dfrac{0.32(1-0.32)}{1200}}\\\\=0.32\pm (1.96)(0.013466)\\\\= 0.32\pm0.02639336=(0.32-0.02639336,\ 0.32+0.02639336)\\\\=(0.29360664,\ 0.34639336)=(29.360664\%,\ 34.639336\%)\approx(29.4\%,\ 34.6\%) [/tex]

Hence, the required confidence interval =[tex](29.4\%,\ 34.6\%)[/tex]

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